• Codefroces B. New Skateboard


    B. New Skateboard
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Max wants to buy a new skateboard. He has calculated the amount of money that is needed to buy a new skateboard. He left a calculator on the floor and went to ask some money from his parents. Meanwhile his little brother Yusuf came and started to press the keys randomly. Unfortunately Max has forgotten the number which he had calculated. The only thing he knows is that the number is divisible by 4.

    You are given a string s consisting of digits (the number on the display of the calculator after Yusuf randomly pressed the keys). Your task is to find the number of substrings which are divisible by 4. A substring can start with a zero.

    A substring of a string is a nonempty sequence of consecutive characters.

    For example if string s is 124 then we have four substrings that are divisible by 4: 12, 4, 24 and 124. For the string 04 the answer is three: 0, 4, 04.

    As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use gets/scanf/printf instead of getline/cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java.

    Input

    The only line contains string s (1 ≤ |s| ≤ 3·105). The string s contains only digits from 0 to 9.

    Output

    Print integer a — the number of substrings of the string s that are divisible by 4.

    Note that the answer can be huge, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

    Examples
    Input
    124
    Output
    4
    Input
    04
    Output
    3
    Input
    5810438174
    Output
    9
    个位十位分情况讨论
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #include <vector>
    #include <iomanip>
    #include <cmath>
    #include <ctime>
    #include <map>
    #include <set>
    using namespace std;
    #define lowbit(x) (x&(-x))
    #define max(x,y) (x>y?x:y)
    #define min(x,y) (x<y?x:y)
    #define MAX 100000000000000000
    #define MOD 1000000007
    #define pi acos(-1.0)
    #define ei exp(1)
    #define PI 3.141592653589793238462
    #define INF 0x3f3f3f3f3f
    #define mem(a) (memset(a,0,sizeof(a)))
    typedef long long ll;
    char a[300006];
    int main()
    {
        scanf("%s",&a);
        ll ans=0;
        int len=strlen(a);
        int k=(int)a[0]-48;
        if(k%4==0) ans++;
        for(int i=1;i<len;i++)
        {
            k=(int)a[i]-48;
            if(k%4==0) ans++;
            int s=(int)a[i-1]-48;
            if((s*10+k)%4==0) ans+=i;
        }
        printf("%I64d
    ",ans);
        return 0;
    }
  • 相关阅读:
    HTTP 方法:GET 对比 POST
    js中return的用法
    Javascript:谈谈JS的全局变量跟局部变量
    ajax请求数据之后在已经有的数据前面打对勾的方法
    JS中的call()和apply()方法区别
    聚簇索引与非聚簇索引的区别
    Android开发(27)--TextView单击链接弹出Activity
    Android 4.2启动代码分析(一)
    Android重启应用程序代码
    java.lang.ClassNotFoundException: org.springframework.web.context.ContextLoaderListener
  • 原文地址:https://www.cnblogs.com/shinianhuanniyijuhaojiubujian/p/7230820.html
Copyright © 2020-2023  润新知